In a comment here, Joel David Hamkins said:

...I think perhaps every set-sized open cover of a bounded interval in the surreals has a finite subcover, but there are proper class open covers with no set-sized subcover. (But I have to think more about this to be sure.) – Joel David Hamkins 15 hours ago

What are references (or proofs) for those two things?

I'll provide the second one, modulo the first one.
So: assume any set-size open cover of a closed interval admits a finite subcover.

Let $\omega$ be as usual, in particular $\omega > n$ for all $n \in \mathbb N$. Consider the closed interval $$ [0,1] := \{x \in \mathbf{No}\;:\; 0 \le x \le 1\} . $$ A proper-class open cover can be made up of all open intervals $$ \left]a-\frac{1}{\omega}, a+\frac{1}{\omega}\right[ $$ where $a$ ranges over the proper class $[0,1]$. But of course no finite set of those covers $[0,1]$, since each of them can contain at most one rational number and $[0,1]$ has infinitely many rationals in it.

Thus we are left with:


Let $[0,1]$ be the closed interval in $\mathbf{No}$ defined above. Is it true that any cover of $[0,1]$ by a set of open intervals admits a finite subcover?


2 Answers 2


Suppose $\mathcal{A}$ is a set-size open cover of $[0,1]$ with open intervals. Without loss of generality, $\mathcal{A}$ is closed under overlapping unions (i.e. the union of any two elements of $\mathcal{A}$ that happens to be an open interval is again in $\mathcal{A}$). Let $B$ be the set of right endpoints $b$ of all intervals $(a,b) \in \mathcal{A}$ such that $a < 0$. We must show that $B$ contains a number bigger than $1$.

Let $\gamma$ be a sufficiently large ordinal, at least larger than all the birthdays of endpoints of intervals from $\mathcal{A}$ and larger than $\omega$. Let $c$ be the simplest number such that $b \leq c$ for every $b \in B$ and $c < d$ for every upper bound of $B$ born before time $\gamma$. If $B$ doesn't contain a number bigger than $1$, then certainly $c \leq 1$ since $1$ was born before time $\gamma$. So $c$ must be covered by an interval $(a,b)$ from $\mathcal{A}$. Since $a$ was born before time $\gamma$, it must be that $a < b'$ for some $b' \in B$. But then if $(a',b')$ is an interval from $\mathcal{A}$ with $a' < 0$ then $(a',b') \cup (a,b) = (a',b)$ shows that $b \in B$, which is impossible since $c < b$. From this contradiction, we conclude that $B$ must contain an element greater than $1$.

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    $\begingroup$ This handles set-sized open covers consisting of intervals, and therefore also set-sized open covers consisting of open sets which are each set-sized unions of intervals. It seems likely to me, however, that we may find set-sized open covers consisting of open classes, proper-class unions of open intervals, with no finite subcover. What do you think? $\endgroup$ Sep 24, 2015 at 16:06
  • $\begingroup$ @Joel, I'm wondering too. I thought your example could be coerced into a counterexample for a set-sized union of open classes but that's not obvious. $\endgroup$ Sep 24, 2015 at 17:43
  • $\begingroup$ I think I can do it; I'll write it up and post a bit later... $\endgroup$ Sep 24, 2015 at 18:08
  • $\begingroup$ I posted it in my answer. $\endgroup$ Sep 24, 2015 at 18:57

$\newcommand\No{\text{No}}$ Here is a proof of the second claim, that there is a proper class-sized open cover of the surreal init interval with no set-sized subcover. In particular, there is no finite subcover. (Note that your open cover does admit a set-sized subcover.) The point is that $\No$ is not order-complete with respect to proper class subsets. To see this, build two sequences $$0=a_0<a_1<\cdots<a_\alpha <\cdots\cdots<b_\alpha<\cdots <b_1<b_0=1$$ of length Ord, with no surreal number filling cut between them. Given $a_\alpha< b_\alpha$, we can let $a_{\alpha+1}$ be the average of $a_\alpha$ and $b_\alpha$, and $b_{\alpha+1}$ the average of $a_{\alpha+1}$ and $b_\alpha$. At limits, the limits are not equal, since every set-sized cut in $\No$ is filled with many distinct surreal numbers, and so we can find $a_\lambda<b_\lambda$ in between, specifically $a_\lambda=\{a_\alpha\mid b_\alpha \}_{\alpha<\lambda}$ and $b_\lambda=\{a_\lambda\mid b_\alpha\}_{\alpha<\lambda}$. Note that the birthdays of these points $a_\alpha$ and $b_\alpha$ are strictly increasing as $\alpha$ increases, and furthermore, the interval $(a_\alpha,b_\alpha)$ contains no surreal numbers with earlier birthdays, since $b_\alpha$ is the successor of $a_\alpha$ among the surreals that are born on that day or earlier. It follows that there can be no surreal number $z$ with $a_\alpha<z<b_\alpha$ for all $\alpha$, because it would have to be born at some ordinal stage $\alpha$, and then it would contradict the claim that $(a_{\alpha+1},b_{\alpha+1})$ has no surreals with earlier birthday.

Thus, let $U_\alpha=[0,a_\alpha)\cup(b_\alpha,1]$. This is a proper-class open cover of $[0,1]$, but it has no set-sized subcover, since any set of $U_\alpha$'s would miss out on the later $a_\beta$'s.

Update. Regarding the open cover property, there are some subtle set/class issues here to be found. François has explained that every set-sized open cover of $[0,1]$ consisting of intervals admits a finite subcover, and it follows immediately from this that every set-sized open cover consisting of set-sized unions of intervals also admits a finite subcover, since one may simply take the set of all intervals appearing in any such union as another open cover and reduce to the finite subcover.

But let me show that there are countably many open proper classes $U_n$ for $n\in\omega$, such that $[0,1]\subset\bigcup_n U_n$, but no finitely many $U_n$ cover $[0,1]$. In fact, we will arrange that the $U_n$ are disjoint open classes, and so we will have an open partition into open proper classes.

To see this, we use the fact that cuts of the type I constructed above appear densely in the surreal numbers, that is, proper class cuts $Z$ for which there is a lower sequence $a_\alpha$ converging up to $Z$ and an upper sequence $b_\alpha$ converging down to $Z$, with no surreal number filling this cut. Specifically, the reader may construct countably many such cuts $Z_n$, each with a lower sequence $a^n_\alpha$ converging up to it, and an upper sequence $b^n_\alpha$ converging down to it, so that there is no surreal number $z$ filling the cut, with $a^n_\alpha<z<b^n_\alpha$ for all $\alpha$. We may construct these cuts so that $0<Z_0<Z_1<\cdots <Z_n<\cdots<1$, meaning that the sequences $a^n_\alpha, b^n_\alpha$ are all separated within the surreal unit interval $[0,1]$, and we may assume furthermore that the upper and lower sequences for each such cut do not overlap or interfere with one another.

Now, let $U_n=(Z_n,Z_{n+1})$, meaning the proper class open union $U_n=\bigcup_\alpha (b^n_\alpha,a^{n+1}_\alpha)$, which is a convex class in the surreals, but the endpoints are not realized in $\No$. Adding the bottom and top, let $U_{-1}=(-2,Z_0)\cup(\sup Z_n,2)$, where by this latter interval, we mean the surreal numbers above all the $b^n_0$'s and below $2$. This is an open interval, since no increasing $\omega$-sequence converges.

Since the cuts $Z_n$ are not filled by any surreal number, it follows that $[0,1]\subset \bigcup_n U_n$, and so we have a countable open cover of the unit interval consisting of open proper classes. But there is no finite subcover, since in fact these open classes form a partition of $[0,1]$ into open classes.

So this is a sense in which the surreal numbers are connected, with respect to separations into disjoint open sets, but it is disconnected, if one allows the separating open classes to be proper classes.

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    $\begingroup$ A simpler counterexample for the second half of your answer is just to take any subset $A\subset[0,1]$ and consider the classes of the form $[0,1]\setminus (A\setminus F)$, where $F$ is finite. These are all open (it is easy to see every sub_set_ of $[0,1]$ is closed) and have no finite subcover if $A$ is infinite. $\endgroup$ Sep 24, 2015 at 19:09
  • $\begingroup$ Eric, that is nice! I'll leave my example, however, since it has the added feature of being an open partition, which seems to clarify the set/class issues about connectedness. $\endgroup$ Sep 24, 2015 at 19:18

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